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Copyright of Decibels Lab Pvt. Ltd. Slide 1
3.Heat Transfer Co-efficient Composite Wall
Prepared by Guided By
Krishna S H Suraj S D
Veerendra Chabbi CEO, Decibels Lab Pvt. Ltd.
Research Interns
Decibels Lab Pvt Ltd
Heat and Mass Transfer Lab
lms.decibelslab.com
Copyright of Decibels Lab Pvt. Ltd. Slide 2
Overall heat transfer co-efficient of composite wall
Image from : Roush Industries
Aim : To determine the overall heat transfer co-efficient for heat transfer through composite material consisting of
mild steel, asbestos and brass
Introduction :
1. Conduction between two bodies 2. Composite bodies
lms.decibelslab.com
Copyright of Decibels Lab Pvt. Ltd. Slide 3
Experiment
Image from :
Image from : Thakar Enterprise.
lms.decibelslab.com
Copyright of Decibels Lab Pvt. Ltd. Slide 4
Equations
! = !#$ = !%$ = !&
• Heat balance equation :
• Newton’s law of cooling in convective heat transfer:
! = ' ∗ )* ∗ ∆,-
• Over all thermal conductivity
! = Input heat from heater(W)
.#$ = Thermal conductivity of MS(W/mK)
.%$ = Thermal conductivity of asbestos(W/mK)
.& = Thermal conductivity of brass(W/mK)
/ = C/s Area (m^2)
,#& = Avg of bottom MS plate ℃
,#1 = Avg of Top MS plate ℃
,%1 = Avg of Asbestos plate℃
,&1 = Ambient top brass plate ℃
∆,- = Change in water Temperature ℃
23 = overall heat transfer co-eff (W/'4
℃)
5#$ = Length of MS steel (m)
5%$ = Length of Asbestos (m)
5%$ = Length of Brass (m)
( Neglecting losses )
! = 2 ∗ / ∗ ∆,
2 =
!
/ ∗ (,#$ − ,&)
(9:*;)
2 =
1
5#$
.#$
+
5%$
.%$
+
5&
.&
(;ℎ?@A3;ABCD)
lms.decibelslab.com
Copyright of Decibels Lab Pvt. Ltd. Slide 5
Hand calculation
Given :
!"#= !$# = !& = 0.006 *
+= 0.15 *
."#/
= 115℃
."#1
= 114 ℃
."#3
= 113 ℃
.&5 = 90 ℃
.&7 = 96℃
.&8 = 93℃
.9_;< = 30 ℃
.9_=>? = 50 ℃
*9@ 0.1
AB
#
CD9 = 4.12
AF
AB.A
G"# = 45
G$# = 0.1662
G& = 100
H = 8.24 JKLLM
H = * ∗ CO ∗ (.9QRS
− .9UV
)
H = X ∗ Y ∗ ∆.
X =
H
Y ∗ (.$[B]
− .$[B^
)
= 22.40
Y = _ ∗
+7
4
= 0.017671
Xa =
1
[
!"#
G"#
+
!$#
G$#
+
!&
G&
]
X =
1
0.006
45
+
0.006
0.1662 +
0.006
100
X= 27.52 (W/*7℃)
Overall heat Transfer co-efficient of composite material = 27.5
e
"℃
Results:
lms.decibelslab.com
Copyright of Decibels Lab Pvt. Ltd. Slide 6
Model Overview – Scilab / Xcos Model
lms.decibelslab.com
Copyright of Decibels Lab Pvt. Ltd. Slide 7
Thank you

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HTLAb3compositewall-210709-211213.pdf

  • 1. lms.decibelslab.com Copyright of Decibels Lab Pvt. Ltd. Slide 1 3.Heat Transfer Co-efficient Composite Wall Prepared by Guided By Krishna S H Suraj S D Veerendra Chabbi CEO, Decibels Lab Pvt. Ltd. Research Interns Decibels Lab Pvt Ltd Heat and Mass Transfer Lab
  • 2. lms.decibelslab.com Copyright of Decibels Lab Pvt. Ltd. Slide 2 Overall heat transfer co-efficient of composite wall Image from : Roush Industries Aim : To determine the overall heat transfer co-efficient for heat transfer through composite material consisting of mild steel, asbestos and brass Introduction : 1. Conduction between two bodies 2. Composite bodies
  • 3. lms.decibelslab.com Copyright of Decibels Lab Pvt. Ltd. Slide 3 Experiment Image from : Image from : Thakar Enterprise.
  • 4. lms.decibelslab.com Copyright of Decibels Lab Pvt. Ltd. Slide 4 Equations ! = !#$ = !%$ = !& • Heat balance equation : • Newton’s law of cooling in convective heat transfer: ! = ' ∗ )* ∗ ∆,- • Over all thermal conductivity ! = Input heat from heater(W) .#$ = Thermal conductivity of MS(W/mK) .%$ = Thermal conductivity of asbestos(W/mK) .& = Thermal conductivity of brass(W/mK) / = C/s Area (m^2) ,#& = Avg of bottom MS plate ℃ ,#1 = Avg of Top MS plate ℃ ,%1 = Avg of Asbestos plate℃ ,&1 = Ambient top brass plate ℃ ∆,- = Change in water Temperature ℃ 23 = overall heat transfer co-eff (W/'4 ℃) 5#$ = Length of MS steel (m) 5%$ = Length of Asbestos (m) 5%$ = Length of Brass (m) ( Neglecting losses ) ! = 2 ∗ / ∗ ∆, 2 = ! / ∗ (,#$ − ,&) (9:*;) 2 = 1 5#$ .#$ + 5%$ .%$ + 5& .& (;ℎ?@A3;ABCD)
  • 5. lms.decibelslab.com Copyright of Decibels Lab Pvt. Ltd. Slide 5 Hand calculation Given : !"#= !$# = !& = 0.006 * += 0.15 * ."#/ = 115℃ ."#1 = 114 ℃ ."#3 = 113 ℃ .&5 = 90 ℃ .&7 = 96℃ .&8 = 93℃ .9_;< = 30 ℃ .9_=>? = 50 ℃ *9@ 0.1 AB # CD9 = 4.12 AF AB.A G"# = 45 G$# = 0.1662 G& = 100 H = 8.24 JKLLM H = * ∗ CO ∗ (.9QRS − .9UV ) H = X ∗ Y ∗ ∆. X = H Y ∗ (.$[B] − .$[B^ ) = 22.40 Y = _ ∗ +7 4 = 0.017671 Xa = 1 [ !"# G"# + !$# G$# + !& G& ] X = 1 0.006 45 + 0.006 0.1662 + 0.006 100 X= 27.52 (W/*7℃) Overall heat Transfer co-efficient of composite material = 27.5 e "℃ Results:
  • 6. lms.decibelslab.com Copyright of Decibels Lab Pvt. Ltd. Slide 6 Model Overview – Scilab / Xcos Model
  • 7. lms.decibelslab.com Copyright of Decibels Lab Pvt. Ltd. Slide 7 Thank you